A hollow cylinder, of radius R and mass M, rolls without slipping down a loop-th

Question

A hollow cylinder, of radius R and mass M, rolls without slipping down a loop-the-loop track of radius r. The cylinder starts from rest at a height h above the horizontal section of the track. What is the minimum value of r so that the cylinder remains on the track all the way around the loop?

I start with mgh = 1/2mv^2 + 1/2Iw^2 which becomes mgh = 1/2mv^2 + 1/2(mR^2)(v^2/r^2) but beyond that I don't know what to do.

Explanation / Answer

Before the cylinder begins rolling, it has potential energy mgh. When it is at the top of the loop-the-loop track, the total energy (kinetic + potential) must equal mgh. For the cylinder to stay on the track, though, means that the normal force at the top of the track is 0. At the top of the track, the forces are the normal force downward and gravity downward. So, the force equation looks like mv^2 / r = N + mg, where N = 0. So, then v^2 / r = g. Now, go back to the energy equation. At the top of the loop, the cylinder has translation and rotational kinetic energy. The total kinetic energy is (1/2) mv^2 + (1/2) I w^2, where I is the moment of inertia and w is the speed of angular rotation. For a cylinder rotating about its long axis, the moment of inertia I = mR^2 / 2, and for circular rotation, w = v / R. So, the kinetic energy can be rewritten (1/2) mv^2 + (1/2) (mR^2/2) (v^2 / R^2 ) = (1/2) mv^2 + (1/4) mv^2 = (3/4) mv^2. The height of the track at the top of the loop is 2r. So the potential energy there is 2mgr. Now, you can set up the complete energy equation. mgh = (3/4)mv^2 + 2mgr --->> gh = (3/4) v^2 + 2gr From the force equation earlier, we know that v^2 = rg. Substituting that in... gh = (3/4) rg + 2gr ---->> h = (3/4)r + 2r. So, the minimum h is 2.75r.

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